Are braided commutators primitive elements of a braided Hopf algebra?

Question

Let $H$ be a braided Hopf algebra. The multiplication on $H \otimes H$ is defined by $(a \otimes b)(c \otimes d) = a \Psi(b \otimes c) d$, $a,b,c,d \in H$.

Let $H = T(V)$. There is a algebra map $\Delta: T(V) \to T(V) \otimes T(V)$ such that $\Delta(v) = 1 \otimes v + v \otimes 1$. Therefore \begin{align} \Delta(xy) = 1 \otimes xy + xy \otimes 1 + x \otimes y + x_{(-1)}.y \otimes x_{(0)}. \end{align}

Consider the braided commutator in $T(V)$. We have \begin{align} x \otimes y - \Psi(x \otimes y) = x \otimes y - x_{(-1)}.y \otimes x_{(0)}. \end{align}

I want to check that $x \otimes y - \Psi(x \otimes y)$ is a primitive element or not.

We have \begin{align} & \Delta( x \otimes y - \Psi(x \otimes y) ) \\ & = 1 \otimes xy + xy \otimes 1 + x \otimes y + x_{(-1)}.y \otimes x_{(0)} - 1 \otimes (x_{(-1)}.y) x_{(0)} \\ & \quad - (x_{(-1)}.y) x_{(0)} \otimes 1 - x_{(-1)}.y \otimes x_{(0)} - ( x_{(-1)}.y )_{(-1)}.x_{(0)} \otimes ( x_{(-1)}.y )_{(0)} \\ & = 1 \otimes xy + xy \otimes 1 + x \otimes y - 1 \otimes (x_{(-1)}.y) x_{(0)} \\ & \quad - (x_{(-1)}.y) x_{(0)} \otimes 1 - ( x_{(-1)}.y )_{(-1)}.x_{(0)} \otimes ( x_{(-1)}.y )_{(0)} \end{align}

It seems that $x \otimes y \neq ( x_{(-1)}.y )_{(-1)}.x_{(0)} \otimes ( x_{(-1)}.y )_{(0)}$. Therefore $x \otimes y - \Psi(x \otimes y) $ is not a primitive element of $T(V)$? Thank you very much.

Answer 1

Sidenote: In the situation you describe, $V$ has to be a Yetter--Drinfeld module over $H$, i.e. $$h_{(1)}x_{(-1)}\otimes h_{(2)}.x_{(0)}=(h_{(1)}.x)_{(-1)}h_{(2)}\otimes (h_{(1)}.x)_{(0)}$$ has to be satisfied for all $x\in V$.

It is in general not true that braided commutators are primitive elements in a braided Hopf algebra. To examine the expression you derive more closely, we can see that for degree reasons ($x,y$ are degree one elements), the braided commutators are primitive elements in the braided Hopf algebra if and only if $$x\otimes y=( x_{(-1)}.y )_{(-1)}.x_{(0)} \otimes ( x_{(-1)}.y )_{(0)}.$$ This condition means precisely that $\Psi^2(x\otimes y)=x\otimes y$, which is not true for all elements in a general braided Hopf algebra.

For example, for a diagonal braiding $\Psi(x_i\otimes x_j)=q_{ij}x_{j}\otimes x_i$, this means that $q_{ij}q_{ji}=1$ which is generally not assumed. If there exists a division $I=I_1 \cup I_2$ or the index set $I$ for a diagonal basis of $V$ such that $q_{ij}q_{ji}=1$ for all $i\in I_1$ and $j\in I_2$, then $T(V)$ can be written as a braided tensor product $T(V_1)\underline{\otimes} T(V_2)$ for $V_1$ being the YD-module generated by $x_i$ for $i\in I_1$, and $V_2$ generated by the $x_j$ for $j\in I_2$. (see [AS, 4.2])

So having braided commutators as primitive elements is a special situation.

[AS]: Andruskiewitsch, N. and Schneider, HJ.: Finite quantum groups and Cartan matrices, Adv. in Math. 154 (2000), 1–45.