Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 2 / 5

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.

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.