# Simple quotients of a triple tensor product

Let $$\mathcal{H}$$ be a Hopf algebra over $$\mathbb{C}$$. Let also $$V_1, V_2, V_3$$ finite-dimensional simple modules over $$\mathcal{H}$$ and $$Q$$ be a simple quotient of $$V_1\otimes V_2\otimes V_3$$. Is it possible to show that one of the following statements is true? Is there any counterexample?

i) $$Q$$ is a quotient of $$N\otimes V_3$$, for some simple quotient $$N$$ of $$V_1\otimes V_2$$;

ii) $$Q$$ is a quotient of $$V_1\otimes P$$, for some simple quotient $$P$$ of $$V_2\otimes V_3$$.

Both of these statements are true (at least if $$H$$ is semisimple). It suffices to prove the first one. By hypothesis there is a nonzero map $$V_1 \otimes V_2 \otimes V_3 \to Q$$. It dualizes to a nonzero map $$V_1 \otimes V_2 \to Q \otimes V_3^{\ast}$$ (I don't know if I need to distinguish between left and right duals here if $$H$$ isn't cocommutative but I don't think it matters, just whichever dual makes this true), which factors through its image $$P \to Q \otimes V_3^{\ast}$$. Dualizing again we get a nonzero map $$P \otimes V_3 \to Q$$.
• Thank you very much! However, in this case, if $f: V_1\otimes V_2\rightarrow Q\otimes V_3^*$, then $Im(f)=P$ is not necessarily simple. Am I right? Oct 9, 2020 at 18:00
• @cl4y70n____: ah, you're right, I thought I didn't need semisimplicity as a hypothesis but I forgot that you asked for $Q$ to be simple. I don't know what happens if $H$ isn't semisimple. Oct 9, 2020 at 18:03