Group algebras, finite-group function algebras, and polynomial Hopf algebras

By [F1], $kG$ is the free vector space with basis $G$. The group product on basis elements extends to the bilinear pairing $(\sum a_{g}g,\sum b_{h}h)\mapsto \sum a_g{b}_h(gh)$, with finite output support; [F9] gives its multiplication map on $kG\otimes kG$. Associativity follows on basis triples from $(gh)r=g(hr)$ and then on all finite sums by bilinearity, while $e$ is a two-sided unit. Define $\mathrm{\Delta }$ and $\epsilon$ by their displayed values on this basis and extend linearly.

By [F2] and [F4], the assignments $x\mapsto x\otimes 1+1\otimes x$ and $x\mapsto 0$ extend uniquely to algebra maps $\mathrm{\Delta }:k[x]\to k[x]\otimes k[x]$ and $\epsilon :k[x]\to k$. The induced maps $\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d}$ and $\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta }$ are algebra maps by [F4], [F8], with their codomains identified by the coherent associators [F7]. Both iterated coproduct maps send $x$ to $x\otimes 1\otimes 1+1\otimes x\otimes 1+1\otimes 1\otimes x$ and send $1$ to $1$, so they agree by the universal property [F2]; likewise each counit composite and ${\mathrm{i}\mathrm{d}}_{k[x]}$ agree on $x$ and $1$. Thus $k[x]$ is a bialgebra.

Suppose now that $G$ is finite. The point masses ${\delta }_g$ form a basis of $k^G$, and [F3] identifies the basis ${\delta }_x\otimes {\delta }_y$ with the point-mass basis of $k^{G\times G}$. Thus the formula $\mathrm{\Delta }(f)(x,y)=f(xy)$ defines a linear map into $k^G\otimes k^G$. On $(x,y,z)\in G^3$, the two iterated coproducts evaluate to $f((xy)z)$ and $f(x(yz))$, which agree by associativity in $G$, with the two parenthesizations identified by [F7]; the left and right counit composites evaluate to $f(ex)$ and $f(xe)$, both $f(x)$. Pointwise multiplication and evaluation at $e$ make $k^G$ a unital algebra, and $\mathrm{\Delta }$ and $\epsilon$ preserve products and units pointwise, so the bialgebra compatibility holds.

In the polynomial bialgebra of step 1.2, the bialgebra conditions [F5] give $\mathrm{\Delta }(1)=1\otimes 1$ and $\epsilon (1)=1$, so $1$ is group-like. The formulas for $\mathrm{\Delta }(x)$ and $\epsilon (x)$ in the Statement and [F12] therefore show that $x$ is primitive relative to this unit in every characteristic.

Write $x_g$ for the generator indexed by $g$ in the free associative algebra $k⟨G⟩$ and let $E=\{1-x_e\}\cup \{x_g{x}_h-x_{gh}:g,h\in G\}$, where $gh$ in the subscript is the group product. Set $Q:=k⟨G⟩\mathrm{/}(E)$. The assignment $x_g\mapsto g$ extends to an algebra map $k⟨G⟩\to kG$ by [F10]; it kills $E$ because $e$ is the unit and the multiplication in step 1.1 extends the group law. By [F11] it descends to $\bar{\varphi }:Q\to kG$. Define $\psi :kG\to Q$ linearly by $\psi (g)=[x_g]$. It is unital since $[x_e]=1_Q$, and it preserves basis products because $[x_g][x_h]=[x_{gh}]=\psi (gh)$ by the relations; bilinearity makes it an algebra map. The composite $\bar{\varphi }\psi$ fixes every basis vector of $kG$, so it is the identity. The maps $\psi \bar{\varphi }$ and ${\mathrm{i}\mathrm{d}}_Q$ agree on the generator classes $[x_g]$; uniqueness in [F10]–[F11] gives their equality. Thus the quotient presentation descends to the explicitly constructed group algebra of step 1.1.

Define $S(x^n)=(-1{)}^n{x}^n$ linearly. Put $X=x\otimes 1$ and $Y=1\otimes x$; they commute in the tensor-product algebra, and $\mathrm{\Delta }(x^n)=(X+Y{)}^n$ because $\mathrm{\Delta }$ is an algebra map. For commuting $X,Y$, induction gives $(X+Y{)}^n={\sum }_{j=0}^n\left(\genfrac{}{}{0px}{}{n}{j}\right)X^j{Y}^{n-j}$: the case $n=0$ is $1$, and multiplication by $X+Y$ proves the next case by collecting coefficients and using Pascal's identity $\left(\genfrac{}{}{0px}{}{n}{j}\right)+\left(\genfrac{}{}{0px}{}{n}{j-1}\right)=\left(\genfrac{}{}{0px}{}{n+1}{j}\right)$ (which follows from the factorial formula, with out-of-range binomial coefficients set to zero). Thus $\mathrm{\Delta }(x^n)={\sum }_{j=0}^n\left(\genfrac{}{}{0px}{}{n}{j}\right)x^j\otimes x^{n-j}$. Both antipode products on $x^n$ equal $x^n{\sum }_{j=0}^n\left(\genfrac{}{}{0px}{}{n}{j}\right)(-1{)}^j=(1-1{)}^n{x}^n$, which is $1$ for $n=0$ and $0$ for $n>0$; these are exactly $u\epsilon (x^n)$. Monomials form a basis, so linearity proves both equations for every polynomial, in every characteristic. This proves the Hopf assertion by [F6].

Define $S(f)(x)=f(x^{-1})$. Evaluating the two antipode products at $x\in G$ gives $\mathrm{\Delta }(f)(x^{-1},x)=f(e)$ and $\mathrm{\Delta }(f)(x,x^{-1})=f(e)$: evaluation after $m(S\otimes \mathrm{i}\mathrm{d})$ or $m(\mathrm{i}\mathrm{d}\otimes S)$ is respectively evaluation of the function on $G\times G$ at $(x^{-1},x)$ or $(x,x^{-1})$. Both are the constant function $u\epsilon (f)$, proving the two equations and the Hopf assertion by [F6].

Let $\mathrm{char}k=p>0$. The integer $p$ is prime: if $p=ab$ with $1<a,b<p$, minimality of the characteristic makes $a{1}_k$ and $b{1}_k$ nonzero, but their product is $p{1}_k=0$, impossible in a field. For $0<j<p$, the factorial formula gives $j\left(\genfrac{}{}{0px}{}{p}{j}\right)=p\left(\genfrac{}{}{0px}{}{p-1}{j-1}\right)$; since $j$ is invertible modulo the prime $p$, reducing this identity modulo $p$ shows $p$ divides $\left(\genfrac{}{}{0px}{}{p}{j}\right)$. By step 2.3, these intermediate coefficients vanish in $k$, while the endpoint terms give $\mathrm{\Delta }(x^p)=x^p\otimes 1+1\otimes x^p$. Since $\epsilon$ is an algebra map in this bialgebra and $\epsilon (x)=0$, $\epsilon (x^p)=\epsilon (x{)}^p=0$. By step 2.1 and [F12], $x^p$ is primitive relative to the group-like unit $1$.

For each basis element $g$, both iterated coproducts equal $g\otimes g\otimes g$ under the coherent associator [F7], and both counit contractions equal $g$. On basis pairs $g,h$, $\mathrm{\Delta }(gh)=gh\otimes gh=(g\otimes g)(h\otimes h)$ and $\epsilon (gh)=1=\epsilon (g)\epsilon (h)$; also $\mathrm{\Delta }(e)=e\otimes e$ and $\epsilon (e)=1$. Bilinear or linear extension gives the bialgebra axioms of [F5] on all of $kG$.

Extend $S(g)=g^{-1}$ linearly. On each basis element, $m(S\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(g)=g^{-1}g=e=u\epsilon (g)$ and $m(\mathrm{i}\mathrm{d}\otimes S)\mathrm{\Delta }(g)=g{g}^{-1}=e=u\epsilon (g)$. Each side is linear in $g$, so the two equations hold for every finite sum, proving the Hopf assertion by [F6].

Steps 1.1, 2.2, 3.2 and 4.1 verify the group-algebra structure, its quotient presentation, bialgebra equations and antipode; steps 1.2, 2.1, 2.3 and 3.1 verify the polynomial bialgebra, its characteristic-dependent primitive elements and antipode; steps 1.3 and 2.4 verify the finite-group function bialgebra and antipode. The finite group hypothesis is used precisely for $k^G\otimes k^G\cong k^{G\times G}$ in step 1.3; no such identification is asserted for infinite $G$.