Does the canonical element associated to a finite dimensional $\mathbb{C}^* $-Hopf algebra always have finite order?

Question

Let $\mathcal{A}$ be a finite dimensional $\mathbb{C}^* $-Hopf algebra. Let $B(\mathcal{A})$ be a basis of $\mathcal{A}$ and let $B(\mathcal{A}^* )=\{ \delta_x\in\mathcal{A}^* | x\in B(\mathcal{A}) \}$ be the corresponding dual basis of the dual Hopf algebra $\mathcal{A}^* $ satisfying $\delta_x(y)=\delta_{x,y}$. The canonical element of $\mathcal{A}\otimes\mathcal{A}^* $ defined as $$ \mathbf{c}=\sum_{x\in B(\mathcal{A})} x\otimes \delta_x\in \mathcal{A}\otimes\mathcal{A}^* . $$ It is straightforward to show that $\mathbf{c}$ is an invertible element of $\mathcal{A}\otimes\mathcal{A}^* $ and its inverse is given by $$\mathbf{c}^{-1}=(S\otimes \mathrm{id})\mathbf{c}=\sum_{x\in B(\mathcal{A})} S(x)\otimes \delta_x,$$ where $\mathrm{id}(x):=x$ is the identity map. The antipode axiom directly leads to $$\mathbf{c}^{-1}\cdot\mathbf{c}=\mathbf{c}\cdot\mathbf{c}^{-1}=1\otimes \epsilon,$$ where $1\otimes \epsilon$ is the unit element of $\mathcal{A}\otimes\mathcal{A}^* $.

Furthermore, the $* $-structure in $\mathcal{A}$ implies that $\mathbf{c}^{-1}=\mathbf{c}^* $, i.e., $\mathbf{c}$ is a unitary element of the algebra $\mathcal{A}\otimes\mathcal{A}^* $.

Conjecture: for any finite dimensional $\mathbb{C}^* $-Hopf algebra $\mathcal{A}$, $\mathbf{c}$ has finite order in $\mathcal{A}\otimes\mathcal{A}^* $, i.e., there exists a positive integer $n$ such that $$\mathbf{c}^{n}=1\otimes \epsilon,$$ or equivalently, using Sweedler notation, $$ \sum_{(z)} z_{(1)}z_{(2)}\ldots z_{(n)}=\epsilon(z)1,\quad\forall z\in \mathcal{A}. $$

This conjecture is easily verified to be true for the group algebra $\mathbb{C}[G]$ of any finite group $G$, since there must exist an integer $n$ such that $g^n=1,~\forall g\in G$. I also verified this conjecture computationally in several $\mathbb{C}^* $-Hopf algebras of small dimension. But is it true in general and how to prove it?

Answer 1

Your equation $$\sum z_{(1)} z_{(2)} \cdots z_{(n)} = \epsilon(z) 1$$ is the equation that Kashina uses to define the exponent of a Hopf Algebra. ("A generalized power map for Hopf algebras", Kashina, 2000.) According to Theorem 4.1 in On the Exponent of Finite-Dimensional Non-Cosemisimple Hopf Algebras, if $A$ is a non-cosemisimple Hopf algebra over a field of characteristic zero, whose co-radical is a finite dimensional Hopf-subalgebra, then the exponent of $A$ is infinite.

Unfortunately, this paper does not give any concrete example of such a Hopf algebra, and I am not familiar enough with the field to construct one.

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As the OP has now explained, the $C^{\ast}$ condition is enough to answer the question. I'm not familiar with $C^{\ast}$ Hopf algebras. However, I now understand enough of Etingof and Gelaki to understand their Remark 2.4, about how to construct a finite dimensional Hopf algebra with infinite exponent, so I thought I would explain it here.

Let $n$ be an integer $\geq 2$ and let $q$ be a primitive $n$-th root of unity. (If you like, you can just take $n=2$ and $q=-1$.) Let $A$ be the algebra $$\mathbb{C} \langle g,x \rangle / \langle g^n=1,\ x^n=0,\ gx=qxg \rangle.$$ Clearly, this is an $n^2$ dimensional vector space with basis $\{ x^a g^b : 0 \leq a,b \leq n-1 \}$.

Define $\Delta : A \to A \otimes A$ as the algebra homomorphism with $$\Delta(g) = g \otimes g,\ \Delta(x) = x \otimes 1 + g \otimes x.$$ One must check that this obeys the defining relations of $A$, in other words, that $(g \otimes g)^n = 1 \otimes 1$, $\left( x \otimes 1 + g \otimes x \right)^n = 0$ and $(g \otimes g)\left( x \otimes 1 + g \otimes x \right) = q \left( x \otimes 1 + g \otimes x \right) (g \otimes g)$. The first and last equations are easy.

For the second equation, group together all the terms which use $k$ of the $(x \otimes 1)$'s and $n-k$ of the $(g \otimes x)$'s. If $k=0$ or $k=n$, this is clearly $0$, so suppose that $0<k<n$.

Each term is $s \otimes x^{n-k}$, where $s$ is the shuffle of $n-k$ $g$'s and $k$ $x$'s. It is equal to $\zeta^{i_1+i_2+\cdots+i_{n-k}} x^k g^{n-k}$, where the $i_j$'s are the positions of the $g$'s. Summing as $\{i_1, i_2, \ldots, i_{n-k} \}$ ranges over all $n-k$ element subsets of $\{ 1,2,\ldots, n \}$, we get the coefficient of $T^k$ in $\prod_{i=1}^n (T+\zeta^i)$. That coefficient is $0$. This completes the verification that $\Delta$ is an algebra homomorphism.

To check that $\Delta$ is co-associative, just check that $(\Delta \otimes 1) (\Delta(g)) = (1 \otimes \Delta) (\Delta(g)) = g \otimes g \otimes g$ and $(\Delta \otimes 1) (\Delta(x)) = (1 \otimes \Delta) (\Delta(x)) = x \otimes 1\otimes 1 + g \otimes x \otimes 1 + g \otimes g \otimes x$.

The antipode is the algebra anti-homomorphism with $S(g) = g^{-1}$ and $S(x) = -g^{-1} x$; details are left to the reader. Observe that $S^2(x) = S(-g^{-1} x) = - S(x) S(g^{-1}) = g^{-1} x g = q^{-1} x$, so this Hopf algebra is not involutive.

The co-unit is $\epsilon(g)=1$, $\epsilon(x)=0$; details are likewise left to the reader.

Then $$\Delta^{[m]}(x) = \sum_{i=0}^m g^{\otimes i} \otimes x \otimes 1^{\otimes(m-i)}$$ and multiplying this gives $$\sum_{i=0}^m g^i x = \sum_{i=0}^m g^{i \bmod n} x$$ where $i \bmod n$ is the integer congruent to $i$ modulo $n$ lying between $0$ and $n-1$. Since $\{ g^j x : 0 \leq j \leq n-1 \}$ are linearly independent, this never equals $\epsilon(x) = 0$.

Answer 2

As mentioned in David's answer, the smallest such $n$ is called the exponent of the Hopf algebra $H$. It was first conjectured by Kashina that that the exponent of a finite dimensional semisimple Hopf algebra $H$ is finite , and it is proved in Etingof & Gelaki '99, which further shows that $n$ divides $\dim(H)^3$, and is equal to the order of the Drinfeld element $u$ of the quantum double $D(H)$. The stronger conjecture that $n$ divides $\dim(H)$ is proved for triangular semisimple Hopf algebras in characteristic zero, but appears open for more general semisimple Hopf algebras. The generalization to weak Hopf algebras appears open.