Weak integral forms and the sixth Kaplansky conjecture

TL;DR

This paper proves that finite dimensional semisimple Hopf algebras with a weak integral form are of Frobenius type, connecting integral forms to algebraic structure in the context of the sixth Kaplansky conjecture.

Contribution

It explicitly states and proves a folklore result linking weak integral forms to Frobenius type in semisimple Hopf algebras, using a historical argument.

Findings

Finite dimensional semisimple Hopf algebras with weak integral forms are of Frobenius type.

The proof uses an argument similar to Fossum's, predating the Kaplansky conjectures.

The note clarifies a previously unpublished result referenced by Cuadra and Meir.

Abstract

It is a short unpublished note from 1998. I make it public because Cuadra and Meir refer to it in their paper. We precisely state and prove a folklore result that if a finite dimensional semisimple Hopf algebra admits a weak integral form then it is of Frobenius type. We use an argument similar to that of Fossum \cite{fos}, which predates the Kaplansky conjectures.

Equations

$$\chi (z)=\frac{k}{n}\sum _{(\Lambda )}\chi ({\Lambda }_1)\chi (S({\Lambda }_2))=k⟨\chi ,\chi ⟩=k,$$

$$\eta (z)=\frac{k}{n}\sum _{(\Lambda )}\chi ({\Lambda }_1)\eta (S({\Lambda }_2))=k⟨\chi ,\eta ⟩=0.$$

$$\sum _{(h)}S(h_1)\Lambda h_2=\epsilon (h)\Lambda ;$$

$$zh=\sum _{(h)}{h}_{3}S(h_2)z{h}_1=\sum _{(h)}{h}_2\epsilon (h_1)z=hz.\square$$

$$\frac{n}{k}\text{mbox}{Id}_V=\frac{n}{k}\rho (e)=\chi \otimes \rho (\nu )={\nu }^{i,j}\chi (x_i)\rho (x_j)$$

Full text

We precisely state and prove a folklore result that if a finite dimensional semisimple Hopf algebra admits a weak integral form then it is of Frobenius type. We use an argument similar to that of Fossum [2], which predates the Kaplansky conjectures.

The research was partially supported by NSF

Let $H$ be a semisimple Hopf algebra of dimension $n$ over an algebraically closed field $K$ of zero characteristic. We say that $H$ is of Frobenius type if the degree of of any irreducible representation of $H$ divides $n$. The sixth Kaplansky conjecture [3] is that no Hopf algebra not of Frobenius type exists.

The Hopf algebra operations on $H$ are denoted $m$, $u$, $\Delta$, $\varepsilon$ and $S$. Let us choose $\Lambda\in\int_{H}$ such that $\varepsilon(\Lambda)=n$. We define $\nu\in H\otimes_{K}H$ by $\nu=\sum_{(\Lambda)}\Lambda_{1}\otimes S(\Lambda_{2})$. We fix a simple $H$-module $M$ of dimension $k$ from now on. The representation is $\rho:H\rightarrow\mbox{End}_{K}M$. Its character is $\chi=\mbox{Tr}\circ\rho:H\rightarrow K$. We denote the indecomposable central idempotent corresponding to $M$ by $e$.

Proof. Let $z=\frac{k}{n}\sum_{(\Lambda)}\chi(\Lambda_{1})S(\Lambda_{2})$. Using Larson’s orthogonality relations [4], we see that

[TABLE]

while for a different irreducible character $\eta$

[TABLE]

Thus, it suffices to show that $z$ belongs to the center of $H$. Pick $h\in H$. We recall that $S^{2}=\mbox{Id}_{H}$ by the Larson-Radford theorem [5]. We also note that $\chi(ab)=\chi(ba)$ because of the similar trace property. We start with the equality

[TABLE]

This implies that $z$ is central since

[TABLE]

Let $R$ be a subring of $K$. By a weak $R$-form of $H$, we understand an $R$-form of $(H,m,\nu)$. This is a free $R$-submodule $H_{R}\subseteq H$ with an $R$-basis $x_{1},\,\ldots\,,x_{n}$ that is a $K$-basis of $H$ and the coefficients $m_{i,j}^{k}$ and $\nu^{i,j}$ belong to $R$. These coefficients appear if one writes down $m$ and $\nu$ in terms of the basis, i.e. $m(x_{i}\otimes x_{j})=m_{i,j}^{k}x_{k}$ and $\nu=\nu^{i,j}x_{i}\otimes x_{j}$ keeping summation by the matching sub- and super-scripts in mind. The proof of the following theorem is based on the idea of [2].

Proof. Pick $x\in M$. Let $N=H_{R}\cdot x$. The $R$-module $N$ is torsion-free and, therefore, free since $R$ is a UFD. Let $e_{1},\,\ldots\,,e_{k}$ be an $R$-basis of $N$. Since $N$ is simple, it is also a $K$-basis of $M$.

We can write elements $\rho(t)$ with $t\in H$ as matrices in this basis. If $t\in H_{R}$ then all coefficients of $\rho(t)$ belong to $R$. In particular, $\chi(t)$ belongs to $R$. By Lemma 1,

[TABLE]

The right part is apriori a matrix with coefficients in $R$. Thus, $\frac{n}{k}\in R$ as a coefficient of the left part. $\ \ \Box$

Let ${\mathbb{A}}\subseteq\overline{{\mathbb{Q}}}\subseteq K$ be the ring of integer algebraic numbers. The following corollary is straightforward.

Proof. It suffices to consider the collection of discrete valuation rings of $L$ for all $p$-adic valuations and to apply Corollary 3. $\ \Box$