Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 8 / 5 / Solution

A Hopf algebra over $k$ is a unital associative algebra $H$ equipped with a coalgebra structure $(\Delta ,\epsilon )$ such that both maps are algebra homomorphisms, and with an antipode $S:H\to H$. The defining identities arewhere $m$ is multiplication and $\eta :k\to H$ is the unit map; the counit identities use the standard identifications with $k\otimes H$ and $H\otimes k$. The first compatibility conditions define a bialgebra; the antipode provides its convolution inverse. Multiplication commutativity and coproduct cocommutativity are independent conditions, as the two examples below demonstrate.

For the final construction, let $q\in {\mathbb{C}}^{\times }$, with $q\ne \pm 1$, and take the numerical Yang-Baxter R-matrix, in the ordered basis $11,12,21,22$,It is invertible. Its four action rules are $11\mapsto q11$, $12\mapsto 12+(q-q^{-1})21$, $21\mapsto 21$, $22\mapsto q22$. Applying these rules on the eight basis vectors of $V^{\otimes 3}$ verifies $R_{12}{R}_{13}{R}_{23}=R_{23}{R}_{13}{R}_{12}$, the Yang-Baxter equation.

In the RTT construction, introduce $T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and impose $R{T}_1{T}_2=T_2{T}_{1}R$, where $T_1=T\otimes I$, $T_2=I\otimes T$. Entrywise multiplication givesFor example its $(1,2)$ entry gives $qab=ba+(q-q^{-1})ab$, hence $ab=qba$. DefineThese maps descend to the RTT algebra. Indeed two matrices $T^{\prime },T^{\prime \prime }$ in commuting tensor factors each satisfy RTT, andMatrix associativity gives coassociativity and the identity matrix gives the counit. Thus RTT first gives a bialgebra; the R-matrix alone does not automatically provide an antipode.

Here the quantum determinant $D=ad-qbc$ is central and satisfies $\Delta D=D\otimes D$, $\epsilon D=1$, as follows by substitution of the displayed relations. Impose $D=1$. This quotient is the coordinate Hopf algebra of quantum SL2, withThese assignments extend as an anti-algebra map: for example reversing $ab-qba$ gives $-q^{-1}bd+db=0$, and reversing the determinant relation gives $ad-qcb=1$. The remaining relations are checked in the same way. Direct multiplication gives $TS(T)=S(T)T=I$: the first diagonal entry of $TS(T)$ is $ad-qbc=1$, its second is $da-q^{-1}cb=1$, and the off-diagonal entries vanish by the $q$-commutation relations. These are precisely the antipode identities on generators, proving that the quotient is a Hopf algebra.

To verify both required failures of symmetry, map this algebra onto the quantum Laurent plane with generators $u,u^{-1},v$ and relation $uv=qvu$, by $a\mapsto u$, $b\mapsto v$, $c\mapsto 0$, $d\mapsto u^{-1}$. Construct this algebra directly on the vector space with basis $u^i{v}^j$, $i\in \mathbb{Z}$, $j\ge 0$, and product $(u^i{v}^j)(u^k{v}^{\ell })=q^{-jk}{u}^{i+k}{v}^{j+\ell }$. The exponent identity $-jk-(j+\ell )m=-\ell m-j(k+m)$ makes multiplication associative, and the basis products give $uv=qvu$ with $u$ invertible. Since $uv\ne vu$ for $q\ne 1$, the Hopf algebra is noncommutative. Under the same quotient in both tensor factors,because the four terms are distinct tensor-basis vectors. Hence it is also noncocommutative. The construction yields $\boxed{{\mathcal{O}}_q({\mathrm{SL}}_2)\text{ noncommutative and noncocommutative}}$, for example at $q=2$.