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?
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4$\begingroup$ Probably clear to most readers, but this is false in positive characteristic. A counterexample is $\operatorname{SL}(2,p)$ with $p$ odd as explained in Curtis-Reiner, (17.17). $\endgroup$– Julian KuelshammerCommented Mar 16, 2015 at 10:25
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1$\begingroup$ Though warren was only briefly present here in 2012, it's still worth pointing out that a quick Google Scholar search for "Kaplansky's conjectures" would point to most of the sources mentioned in answers. $\endgroup$– Jim HumphreysCommented Mar 16, 2015 at 13:14
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3 Answers
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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.
answered Sep 29, 2012 at 12:53
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4$\begingroup$ 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. $\endgroup$ Commented Sep 29, 2012 at 16:27
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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 $H$.
answered Sep 29, 2012 at 16:10
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1$\begingroup$ Since some of Kaplansky's conjectures are still open, it's important to include some information about when the paper was published: ams.org/mathscinet-getitem?mr=1761133 (Note too that the author now teaches at U. South Alabama in Mobile.) $\endgroup$ Commented Mar 16, 2015 at 13:10
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There is a new survey on Kaplansky's sixth conjecture by L. Dai and J. Dong, available on the arxiv. Among other results, it mentions the following (always assuming $\operatorname{char} k=0$):
Special primes:
If a semisimple Hopf algebras has a simple module of dimension $p$, where $p=2$ or $p=3$, then its dimension is divisible by $p$.
Low dimension:
Semisimple Hopf algebras of dimension less than $60$ satisfy Kaplansky's sixth conjecture.
Particular properties:
A semisimple Hopf algebra $H$ satisfies Kaplansky's sixth conjecture if it satisfies one of the following conditions:
It is quasitriangular.
Its characters are central in $H^*.$
- It is semisolvable.
Dividing other invariants:
$H$ a semisimple Hopf algebra. $A$ a transitive $H$-module algebra (e.g. $A=\operatorname{End}_k(V)$ for a simple $H$-module $V$). Then $\dim A$ divides $(\dim V)^2\dim H$.
For references, see the survey mentioned above.
answered Mar 16, 2015 at 12:21