Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 2 (Quantum Groups)

Use $q\in k^{\times }$ and the quantum plane convention $yx=qxy$. In this convention $S{L}_q(2)$ is the coordinate Hopf algebra of quantum SL2 with the catalog parameter replaced by $q^{-1}$. It is generated by $a,b,c,d$, subject toEquivalently the last determinant relation is $da-qbc=1$. The quantum determinant convention is essential: these relations use the inverse of the parameter in the catalog's $ab=qba$ convention.

Writing $T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the comultiplication is matrix multiplication, $\Delta t_{ij}={\sum }_k{t}_{ik}\otimes t_{kj}$. Explicitly,The counit is $\epsilon (a)=\epsilon (d)=1$, $\epsilon (b)=\epsilon (c)=0$. Extend $\Delta$ and $\epsilon$ as unital algebra homomorphisms over a field. Matrix multiplication is associative, giving coassociativity; multiplication by the identity matrix gives the counit laws. The antipode isFor example $(TS(T){)}_{11}=ad-q^{-1}bc=1$ and $(S(T)T{)}_{11}=da-qbc=1$, while the off-diagonal entries vanish by the commutation relations. Thus this is a Hopf algebra, not merely an algebra presentation.

Take the right coaction of the coordinate Hopf algebra of quantum SL2 on the quantum plane to beIt is extended multiplicatively and sends $1$ to $1\otimes 1$. The matrix comultiplication and counit verify the coaction identities on $x,y$. To check that it descends to the quantum plane, expandThe last bracket vanishes because $bc=cb$ and $da-ad=(q-q^{-1})bc$.

For the requested calculation, $ca=qac$ and $yx=qxy$ giveMultiply by $\delta (y)$, put all monomials into the order $x^r{y}^s$, and use $bc=cb$. The answer isA left-coaction convention transposes where the matrix coefficients occur; the displayed answer fixes the right coaction throughout.

An $R$-point means a unital algebra homomorphism over a field $\alpha :k_q[x,y]\to R$, as in points of a noncommutative algebra. It is specified by $X=\alpha (x)$ and $Y=\alpha (y)$ satisfying $YX=qXY$. For $R=\mathbb{C}$ and $q$ not a root of unity, $q\ne 1$, so scalar commutativity gives $(1-q)XY=0$. Hence the scalar points are the union of the two coordinate axes:

For matrix-valued points of a noncommutative algebra, put $H={\mathbb{C}}^n$ and let $H_{\lambda }(X)=\ker (X-\lambda I{)}^n$ be a generalized eigenspace. From $XY=q^{-1}YX$,Therefore $Y{H}_{\lambda }(X)\subseteq H_{q^{-1}\lambda }(X)$, interpreting a missing eigenvalue as the zero space. In particular the linear subspaces $V_X={⨁}_{\lambda \ne 0}{H}_{\lambda }(X)$ and $H_0(X)$ are invariant under both matrices. If $\lambda \ne 0$, the numbers $\lambda ,q^{-1}\lambda ,\dots ,q^{-n}\lambda$ are distinct. A nonzero vector $Y^{n}v$ with $v\in H_{\lambda }(X)$ would force all $n+1$ intervening generalized eigenspaces to be nonzero, impossible in dimension $n$. Thus $Y^n{V}_X=0$: $Y$ is nilpotent on $V_X$, although $X$ is invertible there.

Apply the same argument with $X,Y$ exchanged, using $XY=q^{-1}YX$. The sum $V_Y$ of the nonzero $Y$ generalized eigenspaces is invariant under both matrices, and $X$ is nilpotent on it. Consequently $V_Y\subseteq H_0(X)$. Conversely, since $Y$ is nilpotent on $V_X$, its nonzero generalized eigenspaces all lie in $H_0(X)$. Decompose the restriction of $Y$ to that invariant space into its nonzero and zero generalized eigenspaces:The zero generalized eigenspace of $Y$ is $X$-invariant by the scaling identity, so $U$ is invariant under both matrices. On $U$, both matrices have only zero eigenvalues, hence are nilpotent matrices. This proves the spectral decomposition of a quantum-plane point:Here $V_X,V_Y$ mean sums of generalized eigenspaces, not merely ordinary eigenspaces; that distinction is necessary for non-diagonalizable points.

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

A cobraided bialgebra is a bialgebra $H$ equipped with a convolution-invertible bilinear form $r:H\otimes H\to k$ satisfying the following identities, in Sweedler notation:Convolution invertibility means there is $\bar{r}$ with $\sum r(a_{(1)},b_{(1)})\bar{r}(a_{(2)},b_{(2)})=\epsilon (a)\epsilon (b)$, and likewise with $r,\bar{r}$ interchanged. This is a coquasitriangular structure. On right comodules it induces the braidingThe third identity makes this a comodule map, the first two give its two hexagon identities, and the convolution inverse gives the inverse braiding.

For an invertible braiding $C$ satisfying the braid equation on $V\otimes V$, begin with the free associative algebra over a field on matrix coefficients $t_{ij}$, with $\Delta t_{ij}={\sum }_k{t}_{ik}\otimes t_{kj}$ and $\epsilon (t_{ij})={\delta }_{ij}$. Impose precisely the coefficients ofThis quotient is the FRT bialgebra, or Faddeev-Reshetikhin-Takhtajan construction. The relations form a coideal because composing two coactions that preserve $C$ again preserves $C$; the trivial coaction also preserves it, so the counit descends. Universality follows because any such coaction must satisfy exactly these coefficient relations.

For the given $C$, take $t_{11}=a$, $t_{12}=b$, $t_{21}=c$, $t_{22}=d$. Thus $\delta (e_1)=e_1\otimes a+e_2\otimes c$ and $\delta (e_2)=e_1\otimes b+e_2\otimes d$. The coefficient relations, after cancellation of $q^{-1/2}$, areThese are exactly the defining relations of $M_q(2)$ in the inverse-parameter convention of Question 1, without setting the quantum determinant equal to one. They therefore define mutually inverse algebra homomorphisms: each named generator maps to the corresponding $t_{ij}$, and conversely. The matrix comultiplication and counit are the same, so the isomorphism is also one of bialgebras.

Here is the requested check of $ba=qab$ directly from $C$. Apply colinearity to $e_1\otimes e_2$ and compare the coefficient of $e_1\otimes e_1$. On the left, $C(e_1\otimes e_2)=q^{-1/2}{e}_2\otimes e_1$, giving $q^{-1/2}ba$. On the right, the corresponding term of $\delta (e_1\otimes e_2)$ is $e_1\otimes e_1\otimes ab$, and $C(e_1\otimes e_1)=q^{1/2}{e}_1\otimes e_1$. Hence $q^{-1/2}ba=q^{1/2}ab$, proving the relation. The other coefficient comparisons give the remaining displayed relations; no additional determinant relation is imposed in an FRT bialgebra.

With the right-comodule braiding convention in part (a), the defining formula isConsequently the scalar form on generators is the following table, with row the first argument and column the second:For example, the extra $e_2\otimes e_1$ component of $C(e_2\otimes e_1)$ is $r(b,c)=q^{-1/2}(q-q^{-1})$. The other off-diagonal component gives $r(d,a)=q^{-1/2}$.

For arbitrary elements of the FRT bialgebra, extend this table by bilinearity, the unit rule $r(1,z)=r(z,1)=\epsilon (z)$, andThese are an explicit recursive algorithm for any two words in the generators, so the table determines the entire cobraiding. The braid equation for $C$ ensures the algorithm annihilates the defining coefficient relations in either argument: moving one crossing past two successive crossings gives precisely the two reductions of a quadratic colinearity relation. Thus it descends from the free algebra over a field to the FRT bialgebra. For the convolution inverse, use the inverse of $R=\tau C$, where $\tau$ flips the tensor factors: its coefficients are $\bar{r}(t_{ki},t_{lj})=(R^{-1}{)}_{kl,ij}$. Equivalently, in the flipped coefficient formula use $\tau C^{-1}\tau$, not $C^{-1}$ alone. This gives $\bar{r}(a,a)=\bar{r}(d,d)=q^{-1/2}$, $\bar{r}(a,d)=\bar{r}(d,a)=q^{1/2}$, $\bar{r}(b,c)=-q^{1/2}(q-q^{-1})$, and all other generator values zero. This specifies the coquasitriangular structure completely, not just its values on degree-one elements.

Use the multiplicativity rule for the cobraided bialgebra and $\Delta b=a\otimes b+b\otimes d$:The generator table has $r(a,b)=0$, soHere the tensor notation means the value of the scalar form on the two displayed arguments, not multiplication within $M_q(2)$.

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.

For unital algebras over a field $A,B$, a measuring coalgebra is a coalgebra $C$ together with a linear map $p:C\to {\mathrm{Hom}}_k(A,B)$ such that, in Sweedler notation,Thus a group-like element acts by a unital algebra homomorphism over a field, while skew-primitive elements give twisted product rules.

The universal measuring coalgebra $P(A,B)$ has a universal measuring map $\pi :P(A,B)\to {\mathrm{Hom}}_k(A,B)$ with this property: for every measuring map $p$ from a coalgebra $C$, there is a unique coalgebra homomorphism $f:C\to P(A,B)$ satisfying $\pi f=p$. The uniqueness is among coalgebra homomorphisms; $\pi$ itself need not be an injective map of vector spaces.

The printed input $x^r{x}^s$ is a genuine typo in the PDF as well as in the TeX: $E$ annihilates every power of $x$, whereas the displayed right-hand side is generally nonzero. The intended input is $x^r{y}^s$, and the calculation below proves that corrected identity.

The given comultiplications make $C_q$ a coalgebra, with counit one on $K,K^{-1},I$ and zero on $E,F$. Write $I_0=p(I)$, $K_0=p(K)$ and $E_0=p(E)$, and similarly for $F$. The measuring coalgebra identities areThe group-like elements act multiplicatively. Although the statement omits an explicit specification of $p(I)$, its identity action follows from the other data and the quantum plane relation. Applying $E_0$ to $yx=qxy$ gives $q{x}^2=q{I}_0(x)x$, so cancellation in the noncommutative domain gives $I_0(x)=x$. Applying $F_0$ to the same relation gives $q{y}^2=qy{I}_0(y)$, so $I_0(y)=y$. Thus $I_0$ is the identity on the entire quantum plane.

Now $E_0(x^r)=0$ by induction. Suppose $E_0(y^{s-1})=[s-1]x{y}^{s-2}$. The measuring rule and $y^{s-1}x=q^{s-1}x{y}^{s-1}$ giveThe induction starts with $E_0(y)=x$ and $[1]=1$. A final product rule givesFor $s=0$ the answer is zero. By contrast the literal printed input has $\boxed{p(E)(x^r{x}^s)=0}$; it cannot satisfy the printed right-hand side for $s=1$.

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.