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

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?

• The canonical element of $\mathcal A$ is an element of $\mathcal A \otimes \mathcal A^*$? Commented Jul 13 at 16:02
• @LSpice I edited the question reflecting your comment. Maybe I should call it "the canonical element associated to $\mathcal{A}$" to be precise. Commented Jul 13 at 16:07
• Re, I didn't mean to argue against this if it is the accepted terminology—it wouldn't be the first time hysterical raisins stuck us with slightly imprecise terminology. But, if it is original to you, then I would prefer the later terminology you suggest, unless a Hopf algebra can have at most two representations as $\mathcal A \otimes \mathcal A^*$ (there's the obvious possibility of switching $\mathcal A$ and $\mathcal A^*$, which doesn't affect your construction). Commented Jul 13 at 16:11
• In your computations, may one always take $n$ to be the dimension of $\mathcal{A}$? Commented Jul 13 at 23:08
• @ChrisH I only conjecture the existence of such an $n$ for each Hopf algebra $\mathcal{A}$. You conjecture that $n$ can always be taken to be dim$\mathcal{A}$? While I have quite some confidence in my conjecture based on physical intuition, I have no idea if your stronger conjecture is right or not. In particular, I think that my conjecture can be generalized to weak $\mathbb{C}^*$-Hopf algebras, while your conjecture fails for the famous 13-dimensional Fibonacci weak Hopf algebra, where I know $n=5$. Commented Jul 14 at 21:43

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.

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 $$(g \otimes g)^n = 1 \otimes 1$$ and $$\left( x \otimes 1 + g \otimes x \right)^n = 0$$. The first is easy.

For the second, 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.

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 $$\epsilon(x) = 0$$.

• Many thanks! My question is only about the $\mathbb{C}^*$ case, and a reference in the paper you mentioned (Etingof & Gelaki 99') solved the semisimple case. So the conjecture is true! Commented Jul 14 at 23:05

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.