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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.

Use the comultiplication $\Delta E=E\otimes K+1\otimes E$, $\Delta F=F\otimes 1+K^{-1}\otimes F$, $\Delta K=K\otimes K$ of the quantum enveloping algebra of sl2. Choose a basis $u_0,u_1$ for $V_{1,1}$ with $F{u}_0=u_1$, $E{u}_1=u_0$, and $K{u}_i=q^{1-2i}{u}_i$. Choose a basis $w_0,w_1,w_2$ for $V_{1,2}$ withand all terminal raising or lowering actions zero. This is one of the corrected bases from part (a).

In the tensor product of modules, putThe vector $z_0$ is killed by $E$ and has $K$ eigenvalue $q^3$. Direct application of the comultiplication givesand the string formula gives $E{z}_1=[3]z_0$, $E{z}_2=[2]z_1$, $E{z}_3=z_2$. Also $K{z}_i=q^{3-2i}{z}_i$. Thus $W_3=⟨z_0,z_1,z_2,z_3⟩$ is a submodule isomorphic to $V_{1,3}$.

The other highest vector and its lowered partner areIndeed $E{h}_0=([2]q^2-q^2[2])u_0\otimes w_0=0$, $K{h}_0=q{h}_0$, $K{h}_1=q^{-1}{h}_1$, $E{h}_1=h_0$ and $F{h}_1=0$. Therefore $W_1=⟨h_0,h_1⟩$ is a submodule isomorphic to $V_{1,1}$.

To verify that this is a direct sum, in the $q$ weight space the determinant of the coefficients of $z_1,h_0$ is $-q^2-q^{-1}[2]=-[3]\ne 0$. In the $q^{-1}$ weight space, the coefficient determinant of $z_2,h_1$ is $-q[2]-q^{-2}=-[3]\ne 0$. The extreme weight spaces are spanned by $z_0,z_3$. Hence these six vectors form a basis of the entire tensor product of modules. Each summand is a simple module: projecting a nonzero vector onto one of its distinct $K$ weight spaces and applying the nonzero raising and lowering coefficients recovers its entire basis. Consequently

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

Put $A=k_q[x,y]$. First describe the multiplication of the universal measuring coalgebra $P(A,A)$. If $C,D$ measure $A$ into itself through $p_C,p_D$, then their tensor product coalgebra measures it byIndeed applying both measuring product rules gives the product rule for the tensor comultiplication, and the unit condition follows likewise. Apply universality with $C=D=P(A,A)$: composition induces a coalgebra homomorphism $\mu :P(A,A)\otimes P(A,A)\to P(A,A)$. The identity endomorphism induces the unit $k\to P(A,A)$. Associativity and the unit laws follow by uniqueness because both candidate coalgebra homomorphisms evaluate as the same compositions of endomorphisms. Thus $P(A,A)$ is a bialgebra.

The generator actions from Question 4 satisfy all defining relations of the quantum enveloping algebra of sl2, so they define an action of $U_q$ on $A$. Its skew product rules are exactly the rules from $\Delta E=E\otimes K+1\otimes E$ and $\Delta F=F\otimes 1+K^{-1}\otimes F$. Products of elements satisfying the module algebra rule satisfy it as well: apply the two successive actions and use that $\Delta$ is multiplicative. Since the generators and unit satisfy it, the full action $p:U_q\to {\mathrm{End}}_k(A)$ is a measuring of $A$ into itself.

Universality therefore gives a unique coalgebra homomorphism $\rho :U_q\to P(A,A)$ inducing this action. To prove multiplicativity, compare the two coalgebra homomorphisms from $U_q\otimes U_q$ given by $\rho {\mu }_{U_q}$ and ${\mu }_P(\rho \otimes \rho )$. Both induce $(u\otimes v)(a)=u(va)$, so uniqueness makes them equal. The unit comparison is identical. ConsequentlyThis avoids assuming the universal measuring map is injective: equality is justified by its universal property for coalgebra maps.

Closing an identity strand using compatible evaluation morphisms and coevaluation morphisms gives the categorical dimension in a pivotal or ribbon setting. For the standard two-dimensional quantum enveloping algebra of sl2 module, the weighted contraction has diagonal coefficients $q^{-1},q$, giving quantum dimension $q+q^{-1}=[2{]}_q$. The chosen duality and framing conventions matter.

For $yx=qxy$, the quantum enveloping algebra of sl2 acts by $K(x^r{y}^s)=q^{r-s}{x}^r{y}^s$, $E(x^r{y}^s)=[s{]}_q{x}^{r+1}{y}^{s-1}$ and $F(x^r{y}^s)=[r{]}_q{x}^{r-1}{y}^{s+1}$. Negative exponents here have zero coefficients and are omitted. The quantum integer identity $[r][s+1]-[s][r+1]=[r-s]$ verifies the commutator. These operations preserve total degree and make the quantum plane a module algebra.