Taft-Hopf Algebra has dimension $N^2$?

Question

1. Definition of the Taft-Hopf Algebra
   Let $k$ be a field. Let be $N$ a positive integer such that there exists a primitive $N$-th root of unitiy $\zeta$ over $k$. Denote by $(H, \mu, 1_H)$ the unital, associative algebra over $k$ that is generated by elements $g$ and $x$, subject to the three relations: $g^N=1$, $x^N=0$, $xg=\zeta gx$. One can show that there are algebra maps:

• $\eta: k \rightarrow H; 1 \mapsto 1_H$.
• $\Delta: H \rightarrow H \otimes H; g \mapsto g\otimes g, x \mapsto 1 \otimes x + x \otimes g$.
• $\epsilon: H \rightarrow k; g \mapsto 1, x \mapsto 0$.
• $S: H \rightarrow H^{opp}; g \mapsto g^{-1}, x \mapsto -xg^{-1}$.

Then (one can show) that $(H, \mu, \eta, \Delta, \epsilon, S)$ is a Hopf algebra. We call it the Taft-Hopf Algebra.

2. Question(s)
   

• In my lecture notes it says, for a given $N$ the Taft-Hopf algebra has dimension $N^2$. Why?
• What would be a basis?

Answer 1

This has nothing to do with the Hopf algebra structure.

It follows from the relations $g^N = 1$ that we can define for every element $[n]$ of $\mathbb{Z}/N$ the power $$ g^{[n]} := g^n \,. $$ We show in the following that the elements $g^{[n]} x^m$ with $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$ are a basis of $H$.

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Vector space generating set

We first show that these elements span the algebra $H$ as a vector space. We know from the construction of $H$ that it is spanned as a vector space by the monomials $$ p_1 \dotsm p_r $$ with $r \geq 0$ and $p_i \in \{g, x\}$. It follows from the relation $x g = \zeta g x$ that we can reorder the factors $z_i$ in any such monomial up to some nonzero factor (namely a power of $\zeta$). It follows that $H$ is already spanned by those monomials of the form $$ g^n x^m $$ with $n, m \geq 0$. It follows from the relations $g^N = 1$ and $x^N = 0$ the vector space generators can equivalently be written as $$ g^{[n]} x^m $$ with $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$.

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Linearly independent

To show that these monomials are linearly independent we use a standard trick from representation theory.

We first compute the action of the two algebra generators $g$ and $x$ on our vector space generating set. We find from the relations $xg = \zeta gx$ and $x^N = 0$ that \begin{align*} g \cdot g^{[n]} x^m &= g^{[n+1]} x^m \,, \\ x \cdot g^{[n]} x^m &= \begin{cases} \zeta^n g^{[n]} x^{m+1} & \text{if $0 \leq m \leq N-2$,} \\ 0 & \text{if $m = N-1$,} \end{cases} \end{align*} for all $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$.

Let now $V$ be the free vector space with basis $$ G^{[n]} X^m \qquad \text{with $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$.} $$ We denote the free algebra on the generators $g$ and $x$ by $F$. We can define an $F$-module structure on the vector space $V$ via \begin{align*} g \cdot G^{[n]} X^m &= G^{[n+1]} X^m \,, \\ x \cdot G^{[n]} X^m &= \begin{cases} \zeta^n G^{[n]} X^{m+1} & \text{if $0 \leq m \leq N-2$,} \\ 0 & \text{if $m = N-1$,} \end{cases} \end{align*} for all $[n] \in \mathbb{Z}$ and $m = 0, \dotsc, N-1$.

This module structure is compatible with the relations of $H$ because \begin{align*} xg \cdot G^{[n]} X^m &= x \cdot G^{[n+1]} X^m \\ &= \zeta^{n+1} G^{[n+1]} X^{m+1} \\ &= \zeta^{n+1} g \cdot G^{[n]} X^{m+1} \\ &= \zeta g x \cdot G^{[n]} X^m \end{align*} for all $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-2$, and similarly \begin{align*} xg \cdot G^{[n]} X^{N-1} &= x \cdot G^{[n+1]} X^{N-1} \\ &= 0 \\ &= \zeta g \cdot 0 \\ &= \zeta g x \cdot G^{[n]} X^{N-1} \,, \end{align*} for all $[n] \in \mathbb{Z}/N$, as well as $$ g^N \cdot G^{[n]} X^m = G^{[n+N]} X^m = G^{[n]} X^m $$ for all $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$, and $$ x^N \cdot G^{[n]} X^m = 0 = 0 \cdot G^{[n]} X^m $$ for all $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$.

It follows that the $F$-module struture on $V$ descends to an $H$-module structure on $V$.

For this $H$-module structure on $V$ we now have $$ g^{[n]} x^m \cdot G^{[0]} X^{0} = \zeta^{0 \cdot m} g^{[n]} \cdot G^{[0]} X^m = g^{[n]} \cdot G^{[0]} X^m = G^{[n]} X^m $$ for all $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$. The elements $G^{[n]} X^m$ are linearly independent in $V$ (since they form a basis). It thus follows that the elements $g^{[n]} x^m$ are linearly independent in $H$.

Indeed, if we have some linear combination $$ 0 = \sum_{[n], m} \lambda_{[n], m} g^{[n]} x^m $$ in $H$ then it follows that $$ 0 = \sum_{[n], m} \lambda_{[n], m} g^{[n]} x^m \cdot G^{[0]} X^{0} = \sum_{[n], m} \lambda_{[n], m} G^{[n]} X^m $$ and therefore $\lambda_{[n], m} = 0$ for all $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$.

We have altogether shows that the elements $g^{[n]} x^m$ of $H$ with $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$ are vector space generators for $H$ as well as linearly independent. These elements are hence a basis of $H$, as claimed in the beginning.