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 Hopf algebra over a field $k$ and $V$ a finite dimensional left $H$-module. Then $End_{k}(V)$ is a right $H$-module via $(f\cdot h)(v)=S(h_{1})f(h_{2}\cdot v)$.

We set $Ann(End_{k}(V))$={$h\in H: f\cdot h=0, \forall f\in End_{k}(V)$} and $A=H/Ann(End_{k}(V))$.

Let $I$ be the 1-dimensional subspace generated by $id_{V}$. Then $I$ is a submodule of $End_{k}(V)$. Let $Ann(I)$={$\bar{h}\in A: id_{V}\cdot \bar{h}=0$}.

Is there a sufficient condition of $V$ in order to guarantee that $A$ has an ideal $L$ such that $A=L\oplus Ann(I)$?

If $H$ is a group algebra $kG$, then $End_{k}(V)$ is a right $kG$-module via $(f\cdot g)(v)=g^{-1}f(g\cdot v)$.

Can we answer this question in this case?

share|improve this question
add comment

2 Answers 2

Yes, it is sufficient that $H$ is finite dimensional semisimple. It is not necessary because enveloping and quantum enveloping algebras of simple Lie algebras provide other examples.

Overall, this is equivalent to semisimplicity of the category of finite-dimensional $H$-modules. I do not know what structure properties of $H$ ensure this.

share|improve this answer
1  
No. The counterexample is the universal enveloping algebra H of a complex semisimple Lie algebra. Every finite dimensional left H-module V possesses the property mentioned above. –  sife Jun 6 '11 at 8:10
    
You are right: I did not read the question carefully and thought that your AnnI is in $H$, not $A$... I will edit... –  Bugs Bunny Jun 6 '11 at 8:21
    
If we focus on the representation V, what propersitions does V have to insure that A possesses the property mentioned above? –  sife Jun 6 '11 at 8:42
    
If every finite-dimensional H-module is semisimple, then A possesses the property mentioned above. But this answer is trivial........... –  sife Jun 6 '11 at 8:45
    
Off course, it is trivial but the reformulation is clearer: you are asking for which Hopf algebras the category of finite dimensional reps is semisimple... –  Bugs Bunny Jun 7 '11 at 9:27
add comment

One has that $id_V.h=\epsilon(h)id_V$ for all $h \in H$. Thus $Ann_H(I)=H^+=\{h\in H\;|\; \epsilon(h)=0\}$.

Then the annihialtor inside $A$ is $\pi(H^+)$ where $\pi:H \rightarrow A$ is the canonical projection. A complement of it would be given by the image of the integral of $H$ under $\pi$.

I guess the question is when $\pi(\Lambda)$ is not zero? or in other words when $\Lambda \notin Ann_H(V)$. That is the case if and only if $V^H \neq 0$.

share|improve this answer
    
No!!! $Ann(I)$={$\bar{h}\in A:id_{V}\cdot \bar{h}=0$}. –  sife Jun 6 '11 at 16:50
    
I modified my answer a little bit. –  anonymus Jun 6 '11 at 20:11
add comment

Your Answer

 
discard

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.