How to prove the following results for $H$-bimodule $V$?

Question

Suppose $(H, \cdot, \eta, \Delta,\varepsilon,S)$ be a Hopf algebra, and $V$ be a (finite-dimensional) $H$-bimodule.

Then, how can we prove $\alpha\dot{}h-h.\alpha = 0$ for all $h \in H$ if we have $\alpha \in V$ which satisfy $$\mathrm{S}h_{(1)}.\alpha \dot{} h_{(2)} - \varepsilon(h)\alpha = 0$$ for all $h \in H$, where $.$ is a left $H$-action and $\dot{}$ is a right $H$-action on $V$?

Is it possible? or if it's impossible, is there a known condition that $V$ and $\alpha$ must satisfy?

@Berci Thanks for pointing out. It's an typo. $\alpha$ is correct. — ChoMedit, Nov 10 '21 at 13:46
Can you prove it or find counterexample in the specific case $H$ is a group algebra? — Berci, Nov 10 '21 at 13:49
@Berci Yes, it's just the rewriting of the $g^{-1}.\alpha\dot{}g = \alpha$ for $g \in G$ if $H$ is $kG$. — ChoMedit, Nov 10 '21 at 14:20

score 1 · Answer 1 · answered Nov 10 '21 at 14:27

Okay, so I think I can approach to the proof.

By the condition, we can apply the result on each $h_{(2)}$ and have $$ h_{(1)}.Sh_{(2)}.\alpha \dot{}h_{(3)} = \varepsilon(h_{(2)}) h_{(1)}.\alpha $$

It is equivalent to $$ \alpha\dot{}h = \varepsilon(h_{(1)})\alpha\dot{}h_{(2)} = \varepsilon(h_{(2)}) h_{(1)}\dot{}\alpha = h\dot{}\alpha $$ by the properties of $\varepsilon$ in Hopf algebra $H$.

How to prove the following results for $H$-bimodule $V$?

1 Answers1