# “Nice” bases for finite dimensional semisimple Hopf algebras

Let $$H$$ be a finite dimensional semisimple Hopf algebra over $$\mathbb{C}$$. Can one choose a basis $$\{h_1, \dots, h_n \}$$ of $$H$$, where $$h_1 = 1$$, such that if we write $$\Delta(h_i) = \sum_{1 \leq j,k \leq n} \alpha_{ijk} h_j \otimes h_k$$ for some $$\alpha_{ijk} \in \mathbb{C}$$, then $$\alpha_{ij1} = 0$$ for all $$i,j$$, except $$i=1$$ (since obviously $$\Delta(1) = 1 \otimes 1$$)? That is, if we write $$\Delta(h_i)$$ as the sum of $$n$$ tensors where the second tensorand is a basis vector, is $$h_1$$ the only basis vector with a nonzero summand of the form $$- \otimes 1$$?

If $$H = \mathbb{C}G$$ is a group algebra, then the basis consisting of grouplike elements obviously has this property. If $$G = \mathbb{Z}_p$$ is cyclic of order $$p$$, then $$(\mathbb{C} \mathbb{Z}_p)^*$$ has this property (and I think this is true for any finite abelian group). Many examples I know of have this property (e.g. the Kac-Palyutkin Hopf algebra). However, I can't tell if this is true for $$(\mathbb{C} S_3)^*$$, for example.

In general, if we let $$G(H)$$ denote the set of grouplikes, then it might be sensible to consider a basis consisting of unions of cosets of $$G(H)$$. However, I haven't been able to get this to work.

Given a basis for $$H$$ of the sort you are looking for, let $$V$$ be the linear span of $$h_2, \dots, h_n$$. Then $$V \cap \mathbb C1 = 0$$ and $$\Delta(V) \subset V \otimes V$$. Conversely, given such a $$V$$, any basis for $$V$$, together with 1, will do what you ask.
Carefully dualizing, we are asking if the dual Hopf algebra $$H^*$$ must have a 1-dimensional two sided ideal that is not contained in the kernel of the dual of the unit $$\eta^*: H^* \rightarrow \mathbb C$$.
Let try this out when $$H$$ is the dual of a group algebra, so $$H^* = \mathbb C G$$. Is there such a 1-dimensional ideal? Yes: let $$\bar G = \sum_{g \in G}g \in H^* = \mathbb C G$$. The span of this element is a one dimensional ideal, and the element has augmentation $$|G|$$, not 0.
• I believe I worked this out, although using a different method. Let $T \in H^*$ be an integral for $H$, and choose a basis $\{ h_1, \dots, h_n \}$ for $H$, where $h_1 = 1$ (possible since $H$ is semisimple) and $T(h_i) = 0$ for $i \neq 1$ (which can be achieved by subtracting a scalar multiple of $1$ from each basis element, if necessary). Then, writing $\Delta(h_i)$ as in my question and calculating $\delta_{h_i} * T$ using the fact that $T$ is an integral, you find that this basis has the desired properties. – ren Oct 18 at 16:19
• @Nicholas Kuhn: Cute. I don't know the answer in general, but under the assumption of the OP's question, $H$ is semisimple, which is equivalent to $\epsilon(L)\neq 0$ where $\epsilon$ is the augmentation and $L$ the 1-dim. space of left integrals. So $L$ works. – tj_ Oct 18 at 21:39