3
$\begingroup$

As far as I know, low-dimensional semisimple Hopf algebras are classified (along with non-semisimple ones) up to dimension 60, with the first example of a semisimple Hopf algebra not coming from a finite group in dimension 8.

How well are low-dimensional Hopf superalgebras, that is, $\mathbb{Z}_2$-graded Hopf algebras or Hopf algebra objects internal to $\mathbb{Z}_2-\operatorname{Vect}$ understood? Up to which dimension are they classified? Are there interesting semisimple ones? Has someone worked out the representations? I could find an article on the classification of finite dimensional ones up to dimension 4, but it didn't mention semisimplicity or higher dimensional semisimple examples.

Generally, how well are low-dimensional semisimple braided Hopf algebras (internal to a braided category) understood?

$\endgroup$

2 Answers 2

2
$\begingroup$

If H is a Hopf super algebra then $H\#k[\mathbb Z_2]$ is an ordinary Hopf algebra. So, with some work, one should get the classification up to dim 30 from the classification of ordinary Hopf algebras up to dim 60

$\endgroup$
5
  • 1
    $\begingroup$ What the OP asks: "How well are low-dimensional Hopf superalgebras, that is, $\mathbb Z_2$-graded Hopf algebras or Hopf algebra objects internal to $\mathbb Z_2-\operatorname{Vect}$ understood? Up to which dimension are they classified? Are there interesting semisimple ones? Has someone worked out the representations?" What you answer: "with some work, one should get". $\endgroup$
    – Alex M.
    May 16, 2019 at 14:46
  • 2
    $\begingroup$ Well, is something like "do you know groups having $Z_2$ as semidirect factor, for small cardinal? I only found all groups of order less thanks 60". What I suggested is to look at known classification and look for those having a grouplike element of order 2. This is "some work", ok, but it is something, isn't? $\endgroup$ May 16, 2019 at 14:58
  • 1
    $\begingroup$ Given an ordinary Hopf algebra $H_o$ of dimension $2n$, how many different Hopf super algebras $H$ are there such that $H \# k[\mathbb{Z}_2] \cong H_o$? That's not so easy to answer, I think. $\endgroup$ May 16, 2019 at 16:22
  • 1
    $\begingroup$ I don't known the answer un general either, but I known the procedure. If you have an específic H in front of you (I thinking of being un front of the list of dim H leq60), compute the group-like elements. Most of the Hopf algebras un the classification tables área guven by generators that are group-like and skew primitives, so this part shouldn't be hard. Let g be Auch an element. If H=Ho#Z2 then evaluating g in 1 should hice you an álgebra mapa onto Ho. Noe look if you hace a Hopf subalgebra or not. $\endgroup$ May 16, 2019 at 16:30
  • $\begingroup$ sorry about the spelling! I was on my phone. "área guven" -> are given; Auch an element -> such an element ; should hice you ->should give you. Noe look if you hace --> Now look if you have. Sorry again $\endgroup$ May 16, 2019 at 16:36
2
$\begingroup$

The OP asks more than one different things: the classification of fin dim, semisimple (or not), braided Hopf algebras is still a wide open area (up to my knowledge of course).

The classification of finite dimensional hopf superalgebras (the super- here is to be understood as the simplest case of braided) is another thing. It is still open in general, but a particular case of it has been solved long ago; this has to do with all the finite dimensional (super-)cocommutative hopf superalgebras. These have been classified by means of the following:

Let $\mathcal{H}$ be a finite dimensional, super-cocommutative, hopf superalgebra over an algebraically closed field $k$ of characteristic zero. Then $$\mathcal{H}\cong k[G(\mathcal{H})]\ltimes\bigwedge V$$ where $V$ is the space of primitive elements of $\mathcal{H}$ (regarded as an odd vector space), $\bigwedge V$ is its exterior algebra and $k[G(\mathcal{H})]$ is the group algebra of the the finite group $G(\mathcal{H})$ of the grouplikes of $\mathcal{H}$. In other words, $\mathcal{H}$ is a supergroup algebra.

Remarks:

  1. The above proposition generalizes the corresponding classification for the ungraded case (see for example: About the classification of commutative and of cocommutative, fin. dim. Hopf algebras)
  2. The above proposition, can be seen as a corollary (in the fin dim case) of the super-version of the Cartier-Constant-Milnor-Moore classification theorem, which is actually a classification result of super-cocommutative hopf superalgebras:

    Let $\mathcal{H}$ be a super-cocommutative hopf superalgebra over an algebraically closed field $k$ of char zero. Then we have the hopf superalgebra isomorphism: $$ \mathcal{H}\cong k[G(\mathcal{H})]\ltimes_{\pi} U\big(P(\mathcal{H})\big) $$ where, $k[G(\mathcal{H})]$ is the group algebra of the the group $G(\mathcal{H})$ of the grouplikes of $\mathcal{H}$, $U\big(P(\mathcal{H})\big)$ is the universal enveloping algebra of the lie superalgebra $P(\mathcal{H})$ of the primitive elements of $\mathcal{H}$ and the smash product $\ltimes_{\pi}$ is with respect to the representation of $G(\mathcal{H})$ on $P(\mathcal{H})$ determined by: $\pi:G\to Aut(P)$, $\pi(g)x=gxg^{-1}$, for all $g\in G$, $x\in P$.

    Furthermore, the hopf superalgebra $\mathcal{H}$ is finite dimensional if and only if $G(\mathcal{H})$ is a finite group and $P(\mathcal{H})$ is finite dimensional and purely odd (thus: an abelian lie superalgebra).

    For more details, see: theorem 3.3, p.224, B.Kostant, "Graded manifolds, graded Lie theory and prequantization", Differential Geometrical Methods in Mathematical Physics. Lecture Notes in Mathematics, vol 570, p. 177-306, (1977))

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.