3
$\begingroup$

What is a quasi-Hopf algebra filtration? I know what a Hopf-algebra filtration is, but what is the extra condition needed on the Drinfeld associator $\Phi$? And given such a filtration how does one define the corresponding associator on $\text{gr}H$? I cannot seem to find any reference explaining this, so presumably it should be easy.

However I know some examples of quasi-Hopf algebras that are radically graded, i.e. $H=\text{gr}H$ w.r.t. the Jacobson radical where $\text{gr}\Phi=\Phi$, that is $\Phi$ is itself graded in some sense.

1 Answers 1

1

I am not sure if the following remark answers your question, but if $H$ is a finite dimensional quasi-Hopf algebra and $I$ is the radical of $H$, then if $\Delta(I)\subseteq H\otimes I+I\otimes H$ i.e. if $I$ is a quasi-Hopf ideal, then the filtration of $H$ by powers of $I$, is a quasi-Hopf algebra filtration. Thus, the associated graded algebra $gr(H)$ of $H$ under this filtration has a natural structure of a quasi-Hopf algebra. In this paper, the class of radically graded, complex, finite dimensional quasi-hopf algebras whose radical has prime codimension, is investigated. There are interesting references therein, among which I believe that this one may be particularly helpful to your question.

Hope that helps a bit.

  • 0
    My question is how the associated graded algebra becomes a quasi-hopf algebra. Clearly it's an algebra, I understand how to lift the comultiplication. But what is the Drinfeld associator on $\text{gr}(H)$? – 2017-01-08
  • 0
    Let $\text{gr}H=\bigoplus_{k\geq 0} H[k]$ with $H[k]=I^k/I^{k+1}$ and $I$ the radical. Consider the projection $\pi:H\rightarrow H[0]$. I guess $\pi\otimes \pi\otimes \pi(\Phi)$ is the associator on the radically graded quasi-Hopf algebra $H$ and I guess that's the general way of doing it, but I'm unsure. – 2017-01-09
  • 0
    I guess you mean $\big(\pi\otimes \pi\otimes \pi\big)(\Phi)$. Your idea seems interesting. Heve you checked that this element satisfies the defining relations for the associator in $grH$ ? I'll try to do this as soon as I will find time (I am travelling right now). – 2017-01-09
  • 0
    Since we're projecting onto the degree zero part, the relations are trivially satisfied. Indeed, we can obtain them by considering the degree zero part of the defining relations of $\Phi$. So yes, this associator satisfies the defining relations and is still invertible. Now in all examples I know, $\Phi$ only lives in the degree zero, it would be interesting to see examples where this is not the case, but I guess these are not going to be easy. – 2017-01-10
  • 0
    I agree with your observartion. But I am not sure if there is some more general way of doing it. Now regarding the examples you say, do you mean smt like the quasi-Hopf algebras $H(2)$, $H_{\pm}(8)$, $H(32)$, mentioned in sect. 3 of the second one of the references mentioned in my post? – 2017-01-11
  • 0
    Exactly, and those in other papers of Etingof and Gelaki as well as the corresponding Majid algebras in the dual setting. Almost always the associator comes from a 3-cocycle in the group cohomology of the group-like elements. The group-like elements always live in degree $0$. – 2017-01-11