---
tag: TKTMNJQCHM
kind: example
name: Taft algebra
msc: [16T05]
instance_of: [7XFYP0N84H]
uses: [0J817RTVG6]
sources:
  - url: https://en.wikipedia.org/wiki/Taft_Hopf_algebra
same_as:
  wikidata: Q17046143
  wikipedia: Taft Hopf algebra
---

## Example

Let $n \ge 2$ and let $k$ be a field that contains a [[primitive $n$-th root of unity|0J817RTVG6]] $\omega$.
The *Taft algebra* $T_n(\omega)$ is the $k$-algebra generated by $g$ and $x$ with the relations
$$g^n = 1, \qquad x^n = 0, \qquad x g = \omega\, g x.$$
It has the $k$-basis $g^i x^j$ with $0 \le i, j \le n - 1$, so its dimension is $n^2$.
It is a [[Hopf algebra|7XFYP0N84H]] with
$$\Delta(g) = g \otimes g, \quad \Delta(x) = x \otimes 1 + g \otimes x, \quad \varepsilon(g) = 1, \quad \varepsilon(x) = 0, \quad S(g) = g^{-1}, \quad S(x) = -g^{-1} x.$$
The check that $\Delta$ respects the relations: $\Delta(g)^n = g^n \otimes g^n = 1 \otimes 1$.
For $\Delta(x) \Delta(g) = \omega\, \Delta(g) \Delta(x)$, both sides equal $\omega(gx \otimes g + g^2 \otimes gx)$ after using $xg = \omega gx$.
For $\Delta(x)^n = 0$, put $A = g \otimes x$ and $B = x \otimes 1$; then $BA = \omega AB$, so the $q$-binomial formula with $q = \omega$ gives $(A + B)^n = A^n + B^n$, and $A^n = g^n \otimes x^n = 0$, $B^n = x^n \otimes 1 = 0$.
The antipode check on $x$: $S(x) \cdot 1 + S(g) x = -g^{-1}x + g^{-1}x = 0 = \varepsilon(x)$, and $x S(1) + g S(x) = x - x = 0$.

## Remarks

$T_n(\omega)$ is not commutative, since $xg = \omega gx \ne gx$.
It is not cocommutative, since the flip of $\Delta(x)$ is $1 \otimes x + x \otimes g \ne \Delta(x)$.
The antipode satisfies $S^2(x) = g^{-1} x g = \omega x$, so $S^2 \ne \id$; this shows that the antipode of a finite-dimensional Hopf algebra need not be an involution.
For $n = 2$ and $\omega = -1$ (with $\operatorname{char} k \ne 2$) it is Sweedler's $4$-dimensional Hopf algebra, the smallest Hopf algebra that is neither commutative nor cocommutative.

## Explanations

### L1 human

The Taft algebra is a small, explicit Hopf algebra that is "quantum" in both directions.
It is built from a cyclic group element $g$ of order $n$ and a nilpotent element $x$ that do not commute but $q$-commute.
Moving $x$ past $g$ costs a root of unity.
Because everything is finite and explicit, it is the standard first test case for statements about finite-dimensional Hopf algebras.