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An algebra over a commutative ring that is also a coalgebra, whose comultiplication and counit preserve the multiplication and unit. Equivalently it is a bimonoid in $k$-modules.

With the convention $m\ast \gamma =\gamma \ast n$ and invertible bialgebra scalar cocycle $\gamma$ for $n$, the same coalgebra has $n=\bar{\gamma }\ast m\ast \gamma$, or equivalently $m=\gamma \ast n\ast \bar{\gamma }$. Thus $n(a,b)=\sum \bar{\gamma }(a_{(1)},b_{(1)})m(a_{(2)},b_{(2)})\gamma (a_{(3)},b_{(3)})$. Specifying the equation prevents opposite twist conventions from being conflated.

For a comonad, a coalgebra is a map $a:A\to GA$ satisfying ${ϵ}_{A}a=1_A$ and ${\delta }_{A}a=Gaa$. A morphism $f:(A,a)\to (B,b)$ obeys $bf=Gfa$. This is distinct from the tensor-based notion of a coalgebra in algebra.

A linear subspace $I$ of a coalgebra satisfying the displayed conditions. These are exactly the conditions for the comultiplication and counit to descend to the quotient vector space $C/I$. An ideal that is also a coideal in a bialgebra gives a quotient bialgebra.

A right comodule is an object $M$ with a coaction $\rho :M\to M\otimes C$ satisfying the coassociativity and counit laws. This definition extends from coalgebras to comonoids in a monoidal category.

For a coalgebra $C$ and a unital associative algebra over a commutative ring $A$, maps $f,g:C\to A$ have convolution $f\ast g=m_A(f\otimes g){\Delta }_C$, with unit $j_A{\epsilon }_C$. The two-sided convolution inverse of $1_H$ is the antipode.

An element of a coalgebra satisfying the displayed identities. A group-like element in a Hopf algebra is invertible with inverse its antipode image.

For unital algebras over a field $A,B$, a coalgebra $C$ measures $A$ into $B$ when a linear map $p:C\to {\mathrm{Hom}}_k(A,B)$ satisfies the displayed identity and $p(c)(1_A)=\epsilon (c)1_B$. Group-like elements therefore act by unital algebra homomorphisms over a field.

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.

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 $q=2$.

On the monoidal category of modules over the commutative ring $k$, consider the monad $T=H{\otimes }_k-$ coming from the unit and multiplication of the bialgebra $H$. Its opmonoidal functor structure has comparison mapsHere and below Sweedler notation abbreviates the comultiplication $\Delta h=\sum h_{(1)}\otimes h_{(2)}$. The opmonoidal associativity and unit axioms are the coassociativity and counit laws of the coalgebra. The unit and multiplication of a monad are opmonoidal natural transformations because the bialgebra axioms sayThe Eilenberg-Moore category of this opmonoidal monad is the category of left $H$-modules: a monad-algebra map $H{\otimes }_{k}M\to M$ is exactly a unital associative action.

Applying the preceding construction gives the diagonal bialgebra action and the unit actionThe usual associators and unitors for the tensor product of modules are $H$-linear, and the underlying tensor product is exactly ${\otimes }_k$. The forgetful functor into $k$-modules is strict monoidal. The base need not be a field; the modules need be neither flat modules nor finitely generated modules.

A terminal measuring coalgebra: every measuring map from $C$ factors uniquely through its universal measuring map by a coalgebra homomorphism $C\to P(A,B)$. Composition of universal measurings makes $P(A,A)$ a bialgebra, with the identity map providing its unit.