Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $H$ be a semisimple Hopf algebra. One of the Kaplansky's conjectures states that the dimension of any irreducible $H$-module divides the dimension of $H$.

In which cases the conjecture is known to be true?

share|improve this question

2 Answers 2

From Shlomo Gelaki research statement (which is nice survey, by the way):

We also proved that the dimension of an irreducible representation of a semisimple Hopf algebra H, which is either quasitriangular or cotriangular, divides the dimension of H. This result partially answers a celebrated conjecture of Kaplansky, which is still open.

share|improve this answer
It's worth noting that Etingof-Gelaki's full result here is that the dimension of any representation of the Drinfel'd double divides the dimension of the original Hopf algebra. This gives the result in the quasitriangular case (because quasi-triangularity gives a lift of every rep to a rep of the double), but also tells you quite a bit in the non-quasitriangular case. –  Noah Snyder Sep 29 '12 at 16:27

Yorck Sommerhäuser has a very nice survey about Kaplansky's conjectures. Section 6 is devoted to Kaplansky's 6th conjecture.

In Sommerhäuser's survey it is mentioned that Richmond and Nichols proved that the conjecture is true if the simple module has dimension two:

Theorem (Nichols & Richmond). The dimension of a semisimple Hopf algebra over $\mathbb{C}$ is even if the Hopf algebra has a simple module of dimension 2.

In Sommerhäuser's survey it is also mentioned that Montgomery and Witherspoon proved that Kaplansky's conjecture holds if it holds for a subalgebra.

In this paper

Cohen, Miriam; Gelaki, Shlomo; Westreich, Sara. Hopf algebras. Handbook of algebra. Vol. 4, 173--239, Handb. Algebr., 4, Elsevier/North-Holland, Amsterdam, 2006. MR2523421 (2010j:16076), link

it is written that Kaplansky's conjecture has been proved

  • if $H$ is triangular,

  • if $H$ is semisolvable,

  • if $H$ is cotriangular,

  • if $R(H)$ is central in $H^\*$, where $R(H)$ is the span in $H^\*$ of all the characters on $H$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.