# semisimplicity of maps in braided vector spaces

Let $$V$$ be a finite dimensional braided vector space over $$\mathbb{C}$$. This means that we have a map $$c_{V,V}:V\otimes V\to V\otimes V$$ which gives us an action of the braid group $$B_n$$ on $$V^{\otimes n}$$ by letting the generator $$\sigma_i$$ act on the $$i$$th and $$i+1$$st factors in $$V^{\otimes n}$$: $$\sigma_i\mapsto 1\otimes 1\otimes \cdots\otimes c\otimes\cdots\otimes 1$$ The homomorphism of groups $$p_n:B_n\to S_n$$ sending $$\sigma_i$$ to $$s_i:=(i,i+1)$$ does not split, but there is a canonical set-theoretical lifting $$s_n:S_n\to B_n$$, sending a permutation $$\tau= s_{i_1}\cdots s_{i_k}$$ to $$\sigma_{i_1}\cdots \sigma_{i_k}$$, assuming that $$k$$ is the minimal number for which such an expression of $$\tau$$ exists.

Define now the map $$ш_n:V^{\otimes n}\to V^{\otimes n}$$ $$ш_n = \frac{1}{n!}\sum_{\tau\in S_n} s_n(\tau)$$ for every $$n$$. The map $$ш_n$$ appears in the construction of the Nichols algebra of $$V$$. Indeed, it is known that one of the possible ways to define the Nichols algebra $$B(V)$$ of $$V$$ is by $$B(V)_n = V^{\otimes n}/ Ker(ш_n)$$

My question is the following:

Question: Is it true that $$Ker(ш_n) = Ker(ш_n^2)$$?

This must be true in case we know the operator $$ш_n$$ is semisimple. This happens for example in case the braiding on $$V$$ comes from a super vector space structure.

For a general braided vector space, the case $$n=2$$ is equivalent to the fact that the geometric and the algebraic dimension of the eigenvalue -1 of $$c$$ coincide. Is this still true in more general cases? For example when $$V$$ arises from some modular fusion category, or when the action of $$B_n$$ on $$V^{\otimes n}$$ splits via some finite quotient of $$B_n$$?

I have asked this more general question: lifting of idempotents in group ring But got no answer, so I am asking this more specific one again.