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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?

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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.

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    $\begingroup$ 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. $\endgroup$
    – sife
    Jun 6, 2011 at 8:10
  • $\begingroup$ You are right: I did not read the question carefully and thought that your AnnI is in $H$, not $A$... I will edit... $\endgroup$
    – Bugs Bunny
    Jun 6, 2011 at 8:21
  • $\begingroup$ If we focus on the representation V, what propersitions does V have to insure that A possesses the property mentioned above? $\endgroup$
    – sife
    Jun 6, 2011 at 8:42
  • $\begingroup$ If every finite-dimensional H-module is semisimple, then A possesses the property mentioned above. But this answer is trivial........... $\endgroup$
    – sife
    Jun 6, 2011 at 8:45
  • $\begingroup$ 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... $\endgroup$
    – Bugs Bunny
    Jun 7, 2011 at 9:27
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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 $V^H \neq 0$.

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  • $\begingroup$ No!!! $Ann(I)$={$\bar{h}\in A:id_{V}\cdot \bar{h}=0$}. $\endgroup$
    – sife
    Jun 6, 2011 at 16:50
  • $\begingroup$ I modified my answer a little bit. $\endgroup$
    – anonymus
    Jun 6, 2011 at 20:11

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