The Taft algebra: PBW basis and Hopf structure

Let $T$ be the vector space with basis $e_{i,j}$, $i\in \mathbb{Z}\mathrm{/}N\mathbb{Z}$, $0\le j<N$, and extend bilinearly the rule $e_{i,j}{e}_{a,b}={\zeta }^{-aj}{e}_{i+a,j+b}$ when $j+b<N$, and $0$ otherwise. For a triple of basis elements, if $j+b+d\ge N$, both parenthesizations vanish: if an intermediate degree reaches $N$ it is truncated, and otherwise the final multiplication truncates. If $j+b+d<N$, the scalar in $(e_{i,j}{e}_{a,b})e_{c,d}$ is ${\zeta }^{-aj-c(j+b)}$, while the scalar in $e_{i,j}(e_{a,b}{e}_{c,d})$ is ${\zeta }^{-cb-(a+c)j}$; these exponents are equal, and both outputs are $e_{i+a+c,j+b+d}$. Hence multiplication is associative by bilinearity, and $e_{0,0}$ is its two-sided unit.

Put $g=e_{1,0}$ and $x=e_{0,1}$. The formula gives $g^N=e_{0,0}$, $x^N=0$, $gx=e_{1,1}$, and $xg={\zeta }^{-1}{e}_{1,1}$, so $gx=\zeta xg$. Thus the relations define a surjective algebra map from the presented quotient $Q=k⟨g,x⟩\mathrm{/}(g^N-1,x^N,gx-\zeta xg)$ to $T$, since each $e_{i,j}$ is the image of $g^i{x}^j$, by [F2].

In $Q$, the relation $xg={\zeta }^{-1}gx$ moves every $g$ to the left of every $x$; reducing $g$-powers modulo $N$ and setting any $x$-power at least $N$ to zero shows that the $N^2$ words $g^i{x}^j$, $0\le i,j<N$, span $Q$. Their images in $T$ are its distinct basis vectors $e_{i,j}$, hence are linearly independent in $Q$ as well. Therefore $Q\to T$ is an isomorphism and the claimed PBW basis is valid.

Define $\mathrm{\Delta }$ on the free algebra by the stated generator images. They preserve the relations: $\mathrm{\Delta }(g{)}^N=1$, and with $A=1\otimes x$, $B=x\otimes g$ one has $BA=\zeta AB$ and $\mathrm{\Delta }(x)=A+B$. If $(A+B{)}^n={\sum }_{r=0}^n{C}_{n,r}{A}^{n-r}{B}^r$, then $C_{0,0}=1$ and $C_{n+1,r}={\zeta }^r{C}_{n,r}+C_{n,r-1}$, since $B^{r}A={\zeta }^{r}A{B}^r$. For $0\le n<N$ and $0\le r\le n$, the recurrence gives $C_{n,r}={\prod }_{s=1}^r(1-{\zeta }^{n-r+s})\mathrm{/}(1-{\zeta }^s)$: for $1\le r\le n$, set $P={\prod }_{s=1}^{r-1}(1-{\zeta }^{n-r+1+s})$ and $D={\prod }_{s=1}^r(1-{\zeta }^s)$; then $C_{n,r}=P(1-{\zeta }^{n-r+1})\mathrm{/}D$ and $C_{n,r-1}=P(1-{\zeta }^r)\mathrm{/}D$, so ${\zeta }^r{C}_{n,r}+C_{n,r-1}=P(1-{\zeta }^{n+1})\mathrm{/}D=C_{n+1,r}$. For $n=N$ and $0<r<N$, the same formula has nonzero denominator factors and a numerator factor $1-{\zeta }^N=0$; hence $(A+B{)}^N=A^N+B^N=0$. Also $\mathrm{\Delta }(g)\mathrm{\Delta }(x)=g\otimes gx+gx\otimes g^2=\zeta (g\otimes xg+xg\otimes g^2)=\zeta \mathrm{\Delta }(x)\mathrm{\Delta }(g)$, so $\mathrm{\Delta }(gx-\zeta xg)=0$, and $\mathrm{\Delta }$ descends to $Q$.

Extend $S$ as an anti-algebra map from the free algebra using $S(g)=g^{-1}$ and $S(x)=-x{g}^{-1}$. The relation ideal is killed: $S(g^N-1)=g^{-N}-1=0$; reordering with $g^{-1}x={\zeta }^{-1}x{g}^{-1}$ shows $S(x^N)=(-x{g}^{-1}{)}^N$ is a scalar multiple of $x^N{g}^{-N}=0$; and $S(gx-\zeta xg)=S(x)S(g)-\zeta S(g)S(x)=-x{g}^{-2}+\zeta g^{-1}x{g}^{-1}=0$. Thus $S$ descends to $Q$.

The assignment $\epsilon (g)=1,\epsilon (x)=0$ kills $g^N-1$, $x^N$, and $gx-\zeta xg$, so it descends to an algebra map $Q\to k$. Coassociativity holds on $g$ immediately; on $x$, both iterates equal $1\otimes 1\otimes x+1\otimes x\otimes g+x\otimes g\otimes g$. The two counit equations hold on $g,x$ by the given values. These maps are algebra maps, and the presented algebra is generated by $g,x$, so the identities extend to all of $Q$.

On $g$, both $m(S\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }$ and $m(\mathrm{i}\mathrm{d}\otimes S)\mathrm{\Delta }$ give $1=\epsilon (g)1$. On $x$, the left product is $x+(-x{g}^{-1})g=0$, and the right product is $-x{g}^{-1}+x{g}^{-1}=0=\epsilon (x)1$. For products $ab$ of elements on which both identities hold, anti-multiplicativity of $S$ and multiplicativity of $\mathrm{\Delta }$ give $\sum S(a_1{b}_1)a_2{b}_2=\sum S(b_1)(S(a_1)a_2)b_2=\epsilon (a)\epsilon (b)1$, and $\sum a_1{b}_{1}S(a_2{b}_2)=\sum a_1(b_{1}S(b_2))S(a_2)=\epsilon (a)\epsilon (b)1$. Induction on word length and linearity extend both antipode equations to all of $Q$. Thus the given maps make $Q$ a Hopf algebra, with the interface of [F1].