Graded coalgebras, bialgebras and Hopf algebras over a commutative ring

Let $k$ be a commutative ring and let $H={⨁}_{n\ge 0}{H}_n$ be a nonnegatively graded $k$-module (Nonnegatively graded rings and modules, homogeneous elements, and twists). A graded $k$-algebra structure on $H$ consists of a $k$-bilinear associative multiplication $m:H{\otimes }_{k}H\to H$ and a unit $k$-algebra map $u:k\to H$ such that $m(H_i\otimes H_j)\subseteq H_{i+j}$ and $u(k)\subseteq H_0$ (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

A graded coalgebra structure on $H$ consists of $k$-linear maps $\mathrm{\Delta }:H\to H{\otimes }_{k}H$ and $\epsilon :H\to k$, where $k$ is concentrated in degree $0$, such that $\mathrm{\Delta }$ has degree $0$, $\epsilon (H_n)=0$ for $n>0$, and

$$(\mathrm{id}\otimes \mathrm{\Delta })\mathrm{\Delta }=(\mathrm{\Delta }\otimes \mathrm{id})\mathrm{\Delta },(\epsilon \otimes \mathrm{id})\mathrm{\Delta }=\mathrm{id}=(\mathrm{id}\otimes \epsilon )\mathrm{\Delta },$$

using the canonical identifications $k{\otimes }_{k}H\cong H\cong H{\otimes }_{k}k$ (The tensor product $M{\otimes }_{R}N$ from the additive group underlying the free $\mathbb{Z}$-module on $M\times N$, elementary tensors, and finite tensor sums). Here degree $0$ means $\mathrm{\Delta }(H_n)\subseteq {⨁}_{a+b=n}{H}_a{\otimes }_k{H}_b$.

If $H$ is both a graded algebra and a graded coalgebra and $\mathrm{\Delta }$ and $\epsilon$ are $k$-algebra homomorphisms, then $H$ is a graded bialgebra. Equivalently, the multiplication and unit maps $m$ and $u$ are morphisms of graded coalgebras: on $H{\otimes }_{k}H$ use the tensor-product algebra structure $(a\otimes b)(c\otimes d)=ac\otimes bd$ and the tensor-product coalgebra structure with comultiplication $(\mathrm{id}\otimes \tau \otimes \mathrm{id})(\mathrm{\Delta }\otimes \mathrm{\Delta })$ and counit $\epsilon \otimes \epsilon$, where $\tau (b\otimes c)=c\otimes b$ (The tensor product of $R$-algebras has multiplication $(a\otimes b)(a^{\mathrm{\prime }}\otimes b^{\mathrm{\prime }})=a{a}^{\mathrm{\prime }}\otimes b{b}^{\mathrm{\prime }}$, Symmetry and associativity isomorphisms for tensor products over a commutative ring). Indeed, the coalgebra-map identities for $m$ are exactly $\mathrm{\Delta }(ab)=\mathrm{\Delta }(a)\mathrm{\Delta }(b)$ and $\epsilon (ab)=\epsilon (a)\epsilon (b)$, while the identities for $u$ are exactly $\mathrm{\Delta }(1_H)=1_H\otimes 1_H$ and $\epsilon (1_H)=1$; these are the multiplicativity and unit equations for $\mathrm{\Delta }$ and $\epsilon$. The bialgebra is connected when the unit map restricts to an isomorphism $u:k\stackrel{\sim }{\to }{H}_0$; this is the precise meaning of $H_0=k\cdot 1_H$ used for connected graded bialgebras.

For $f,g\in {\mathrm{Hom}}_k(H,H)$, their convolution product is $f\ast g:=m(f\otimes g)\mathrm{\Delta }$. It is associative: its two iterated products are obtained from $(\mathrm{\Delta }\otimes \mathrm{id})\mathrm{\Delta }$ and $(\mathrm{id}\otimes \mathrm{\Delta })\mathrm{\Delta }$ followed by $m(m\otimes \mathrm{id})$ and $m(\mathrm{id}\otimes m)$, respectively, which agree by coassociativity and associativity. Its unit is $u\epsilon$, since the two counit identities give $(u\epsilon )\ast f=f=f\ast (u\epsilon )$. Thus convolution makes ${\mathrm{Hom}}_k(H,H)$ a unital associative $k$-algebra.

An antipode is a $k$-linear map $S:H\to H$ satisfying

$$m(S\otimes \mathrm{id})\mathrm{\Delta }=u\epsilon =m(\mathrm{id}\otimes S)\mathrm{\Delta };$$

equivalently, $S$ is a two-sided convolution inverse of ${\mathrm{id}}_H$. A graded bialgebra equipped with an antipode is a graded Hopf algebra. If $k$ is a field and $H$ is commutative, forgetting the grading gives the usual commutative Hopf-algebra data: the same multiplication, unit, comultiplication, counit and antipode satisfy the same equations, and the antipode is multiplicative. For the latter claim, $S(1)=1$ follows by evaluating the antipode identity at $1$. Give ${\mathrm{Hom}}_k(H{\otimes }_{k}H,H)$ its convolution product from the tensor-product coalgebra structure and multiplication of $H$. Since $m$ is a coalgebra map, $S\circ m$ is a two-sided convolution inverse of $m$. The map $G=m(S\otimes S)\tau$ is another: for $a,b\in H$, its convolution product with $m$ in either order is $\epsilon (a)\epsilon (b)1$, since the Sweedler components reorder by commutativity and the two antipode identities give $\sum S(a_{(1)})a_{(2)}=\epsilon (a)1=\sum a_{(1)}S(a_{(2)})$ and the corresponding identities for $b$. Uniqueness of a two-sided inverse gives $S(ab)=S(b)S(a)=S(a)S(b)$. No commutativity of $H$, field hypothesis, or choice principle is required for the definitions above.